Charge Calculations in Pyrometallurgical Processes

Smelting It is a unit process similar to roasting, to heat a mixture of ore concentrate above the melting point The objective is to separate the gangue mineral from liquid metal or matte The state of the gangue mineral in case of smelting is liquid which is the main difference between roasting and smelting Inputs – Ore, flux, fuel, air Output – Metal or Matte, slag, off-gas

When metal is separated as sulphide from smelting of ore, it is called Matte smelting e.g. Cu2S and FeS When metal is separated as liquid, it is called reduction smelting e.g. Ironmaking Density of liquid metal or matte is around 5-5.5 g/cm3 Density of slag is around 2.8-3 g/cm3 The additives and fluxes serve to convert the waste or gangue materials in the charge into a low melting point slag which also dissolves the coke ash and removes sulphur

Matte Smelting Advantages of matte smelting • Low melting point of matte so that less amount of thermal energy is required by converting the metal of the ore in the form of sulphide and then extracting the metal e.g. melting point of Cu2S and FeS is around 1000 degrees Celsius

• Cu2S which is contained in the matte, does not require any reducing agent It is converted to oxide by blowing oxygen • Matte smelting is beneficial for extraction of metal from sulphide ore, particularly when sulphide ore is associated with iron sulphide which forms eutectic point with Cu and Ni The grade of the matte is defined as the copper grade of matte A matte of 40 percent means, it has 40% copper, so matte is always given in terms of copper, because it is used to produce copper not iron Slag in matte smelting is mixture of oxides e.g. in smelting of copper ore concentrate the slag may contain SiO2, Al2O3, calcium oxide, FeO, Fe2O3, Fe3O4 The desirable properties of slag are low viscosity, solubility, low melting point

Typical reactions in Cu matte smelting: 6CuO + 4FeS = 3 Cu2 S + 4FeO + SO2 or if the O2 pressure is high 6CuO + 4FeS = 2CuSO4 + 2FeS 2CuSO4 + 2FeS = Cu2 S + 2FeO + 3SO2 Cu2 O + FeS = Cu2 S + FeO

Oxygen has greater affinity for iron than copper: 10Fe2 O3 + FeS = 7Fe3 O4 + SO2 3Fe3 O4 + FeS = 10FeO + SO2 In the ideal condition matte contains only Cu2S and FeS, plus little amount of Fe3O4 if oxygen is dissolved Higher oxides of iron are difficult to remove by the slag Roasting has to be controlled in order to minimize the formation of Fe2O3 or Fe3O4, which may enter the matte during smelting Off-gas consists of SO2, nitrogen, oxygen if excess amount of air is used and sometimes SO3 depending on the reaction If fuel is used, CO and CO2 may also be present depending upon the state of combustion

Flash Smelting Conventionally smelting is carried out in reverberatory furnaces, fired with coal or oil Nowadays reverberatory furnaces are being replaced by flash smelting furnaces that have been developed in recent years

The advantages of flash smelting is that it combines both roasting and smelting, whereas in the reverberatory furnace the ore has to be roasted first and then it is transferred to reverberatory furnace for smelting purposes The reason for this combination is the economical processing of large amount of sulphur dioxide that is created especially in roasting Collecting the concentrated off-gas from flash smelting and converting to H2SO4 is much more feasible Other advantages of flash smelting: • Very fine particles of ore concentrates are injected, so the reaction is extremely rapid and very high temperatures are created • Heat generated is sufficient to carry out the smelting

Examples – In a copper ore, chalcopyrite (CuFeS2) is 34%, pyrite (FeS2) is 30% and SiO2 is 36% a) Determine the % Cu and % gangue in the ore b) What % Fe in the ore concentrate is to be removed to make 40% matte? Consider Cu2S c) If only excess S is eliminated in the ore concentrate, what is the composition of the resulting matte? AWCu= 64, AWFe= 56, AWS= 32 aore = Cu2S + gangue % Cu = 34 * (64/184) = 11.83 % Gangue = 100 – 11.83*(160/128) = 85.21 %, % Cu2S = 14.78 % b–

40 100

11.83%

= 14.78%+%𝐹𝑒𝑆

0.4(14.78+%FeS) = 11.83, % FeS = 14.795 after removal of FeO, % Fe = 14.795*(56/88) = 9.415 Initial % Fe = 34*(56/184)+30*(56/120) = 24.35 % % Fe to be removed = 24.35 – 9.415 = 14.935 % c–

CuFeS2 decomposes according to the reaction 2CuFeS2 = Cu2S + 2FeS + S FeS2 decomposes according to the reaction FeS2 = FeS + S % FeS = 24.35*(88/56) = 38.26 % % 𝐶𝑢 Matte grade = % 𝐶𝑢 𝑆+% 𝐹𝑒𝑆 ∗ 100 = 22.2 % 2

Examples – A copper matte may be represented as mCu2S.nFeS with no fixed values of m and n Calculate m and n for a matte grade of 38 % 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐶𝑢 0.38 = 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐶𝑢2 𝑆 + 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐹𝑒𝑆

=

160𝑚 ∗ 128/160 160𝑚 + 88𝑛

60.8𝑚 + 33.44𝑛 = 128𝑚 𝑚 33.44 = ≈ 0.5 𝑛 67.2 Matte may be represented as Cu2S.2FeS or 2Cu2S.4FeS or 3Cu2S.6FeS

Example – Copper ore is smelted in a reverberatory furnace together with a copper concentrate. The fluxes are pure CaCO3 and iron ore Rational Analysis wt%

Material

Cu2S

FeS2

SiO2

Copper ore

17.5

67.5

15

Copper concentrate

35

25

40

Iron ore

20

Fe2O3

80

Iron ore Copper ore

Slag

Reverberatory furnace Copper concentrate

Matte

Flux: CaCO3 Reaction: CaCO3 = CaO + CO2

Iron ore (80% Fe2O3, 20% SiO2) 1000 kg Copper ore 17.5% Cu2S 15% SiO2 67.5% FeS2 Copper concentrate 35% Cu2S 40% SiO2 25% FeS2

Reverberatory furnace

Slag 35% SiO2 20% CaO 45% FeO Matte 37.5% Cu2S 62.5% FeS

Flux: CaCO3

Reaction: CaCO3 = CaO + CO2 Calculate the quantities of concentrate, iron ore and flux in order to smelt 1000 kg of copper ore and obtain a matte grade of 30% Cu and a slag with the composition 35% SiO2 , 20% CaO, 45% FeO Let X be the quantity of Cu concentrate Let Y be the quantity of matte Let Z by the quantity of iron ore Let U be the quantity of slag Four equations are needed to solve for the four variables X, Y, Z, U

Z kg Iron ore (80% Fe2O3, 20% SiO2) 1000 kg Copper ore 17.5% Cu2S 15% SiO2 67.5% FeS2 X kg Copper concentrate 35% Cu2S 40% SiO2 25% FeS2

Reverberatory furnace

U kg Slag 35% SiO2 20% CaO 45% FeO Y kg Matte 37.5% Cu2S 62.5% FeS

Flux: CaCO3

Cu2S balance: Cu2S(in ore) + Cu2S(in Cu conc.) = Cu2S(in matte) Cu2S(in ore) = 17.5% * 1000 = 175 kg Cu2S(in Cu-conc.) = 0.35 * X Cu2S(in matte) = 37.5% * Y = 0.375 Y Equation 1: 175+ 0.35X = 0.375Y

SiO2 balance: SiO2(Cu ore) + SiO2(Cu conc.) + SiO2(iron ore)= SiO2(slag) 0.15% * 1000 + 0.40X + 0.20Z = 0.35U Equation 2: 0.40X + 0.20Z – 0.35U = -150

Fe Balance: Fe(Cu ore) + Fe(Cu conc.) + Fe(iron ore) = Fe(matte) + Fe(slag) Fe(Cu ore) = 67.5% * 1000 * (56/120) = 315 kg Fe(Cu conc.) = 25% * X * (56/120) = 0.117X kg Fe(iron ore) = 80% * Z * (112/160) = 0.56Z kg Fe(matte) = 62.5% * Y * (65/88) = 0.398Y kg Fe(slag) = 45% * U * (56/72) = 0.35U kg Equation 3: 315+0.117X+0.56Z=0.398Y+0.35U

Z kg Iron ore (80% Fe2O3, 20% SiO2) 1000 kg Copper ore 17.5% Cu2S 15% SiO2 67.5% FeS2 X kg Copper concentrate 35% Cu2S 40% SiO2 25% FeS2

Reverberatory furnace

U kg Slag 35% SiO2 20% CaO 45% FeO Y kg Matte 37.5% Cu2S 62.5% FeS

Flux: CaCO3

Sulphur balance: S(Cu ore) + S(Cu conc.) = S(matte), S(flue gas)≈0 S(Cu ore) = (0.175 * (32/160) * 1000) + (0.675 * (64/120) * 1000) = 395 kg S(Cu conc.) = (0.35 * X * (32/160)) + (0.25 * X * (64/120)) = 0.203X kg S(matte) = (0.375 * (32/160) * Y) + (0.625 * (32/88) * Y) = 0.3023Y kg Equation 4: 395+0.203X = 0.3023Y Equation 1: 175+ 0.35X = 0.375Y Z=589.6 kg Iron ore X=3210.2 kg Cu conc. U=7463.4 kg Slag Equation 2: 0.40X + 0.20Z – 0.35U = -150 Y=3462.7 kg Matte Equation 3: 315+0.117X+0.56Z=0.398Y+0.35U CaO in slag: 20% * 7463.4 = 1492.7 kg CaO,

1492.7 * (100/56) = 2665.5 kg CaCO3

Converting Liquid metal or matte coming from the smelting furnace with impurities is converted to high purity metal in oxidizing environments Either steady air, blown air or blown oxygen are utilized to oxidize the gangue species Gangue oxide minerals are removed with the initially forming slag

Inputs – Pig iron, cast iron for steel converting, Cu-Fe matte for copper converting, flux, air Outputs – Slag, steel or blister copper, off-gas Furnaces used Hearths Puddling furnaces Cementation furnaces Bessemer furnaces Open Hearth furnaces Basic oxygen furnaces Electric arc furnaces

Converting Pig Iron Wrought or worked iron was the main malleable iron used in rails and structures until large scale, commercial production of steel It contained low amount of carbon (0.04 to 0.08%) and was worked by hand into bars and various shapes due to its malleability Slag up to 2% is mixed in its microstructure in the form of fibrous inclusions like wood Pig iron and cast iron were initially converted to wrought iron in hearths in ancient times then in puddling furnaces during 18th century In these processes the charge was heated to melting temperature by burning charcoal and oxidized by air Puddling process involves manually stirring the molten pig iron, which decarburerizes the iron As the iron is stirred, globs of wrought iron are collected into balls by the stirring rod and those are periodically removed by the puddler

Horizontal (lower) and vertical (upper) cross-sections of a single puddling furnace. A. Fireplace grate; B. Firebricks; C. Cross binders; D. Fireplace; E. Work door; F. Hearth; G. Cast iron retaining plates; H. Bridge wall

Commercial production of low carbon, low impurity steel was limited to inefficient and expensive process of adding carbon to carbon-free wrought iron between 17th and 19th centuries The manufacturing process, called cementation process, consisted of heating bars of wrought iron in a furnace in between powdered charcoal layers at about 7000 C for about a week to Carbon slowly diffuses into iron and dissolves in the iron, raising the carbon percentage Steel obtained from this process is called “blister steel” due to the blister-like marks formed on the surface due to the evolved gases during the manufacturing process

Up to 3 tons of coke was burnt for each ton of steel produced The fuel and labor costs resulted in a small scale production of steel that was about 8 times more expensive The Bessemer process reduced the time needed to make steel of this quality to about half an hour while only requiring coke to melt the pig iron initially

The Bessemer process - Henry Bessemer patented the process in 1855 The process is carried on in a large ovoid steel container lined with clay or dolomite The capacity of a converter is from 8 to 30 tons of molten iron The key principle is removal of impurities from the iron by oxidation with air being blown through the molten iron The oxidation process removes impurities such as silicon, manganese, and carbon as oxides These oxides either escape as gas or form a solid slag The oxidation also raises the temperature of the iron mass and keeps it molten The refractory lining of the converter also plays a role in the conversion—the clay lining is used in the acid Bessemer, in which there is low phosphorus in the raw material Dolomite, limestone or magnesite are used when the phosphorus content is high in the basic Bessemer

Once the converter is charged with molten pig iron, a strong thrust of air is blasted across the molten mass for about 20 minutes through tuyeres provided at the bottom of the vessel The conversion process called the "blow" is typically completed in around twenty minutes During this period the progress of the oxidation of the impurities is judged by the appearance of the flame issuing from the mouth of the converter since there is not enough time to make material analyses The blow may be interrupted at certain periods to avoid the oxidation of certain impurities Required amount of flux is added at the beginning of each period to produce the slag of desired composition and amount At the end of the process all traces of the silicon, manganese, carbon, phosphorus and sulphur are oxidized, leaving the converter with pure iron In order to give the steel the desired properties, other impurities can be added to the molten steel when conversion is complete

Steel converter analysis A basic pneumatic steel converter is charged with 25 tons of pig iron containing various impurities In addition to the removal of all of the C, Si, Mn and P, iron equivalent to 5% of the weight of charged iron oxidizes at a constant rate throughout the bessemerizing operation Enough lime is added to obtain a slag containing 35% CaO 2/3 of the carbon in steel oxidizes to CO and 1/3 goes to CO2 Air compressor delivers air at a rate of 500 m3/min for specific periods of time Flux CaO

Flue gas CO, CO2, N2

Pig iron

Bessemer converter

Slag Fe2O3, P2O5, SiO2 MnO, CaO

Air

Steel Ultimate Analysis wt% Material

Fe

C

Si

Mn

P

Pig iron

91

3.5

2

1

2.5

Flux CaO 25 tons Pig iron Bessemer converter 91% Fe 3.5% C Air 2% Si 500 m3/min 1% Mn 2.5% P Calculate the volume of air required for the operation Basis 25 tons of pig iron Oxidation step 1: Si + O2 = SiO2 Weight of Si in Pig iron = 0.02 * 25000 = 500 kg nSi= 500/28 = 17.857 kg-atom nO2= 17.857 kg-mole

Flue gas CO, CO2, N2 Slag Fe2O3, P2O5, SiO2 MnO, 35% CaO Steel

Oxidation step 3: C + 1/2O2 = CO, C + O2 = CO2 Weight of C in iron = 0.035 * 25000 = 878 kg nC = 878/12 = 72.917 kg-atom nC(for CO) = (2/3) * 72.917 = 48.611 kg-atom nO2 = (½) * 48.611 = 24.306 kg-mole Oxidation step 2: 2Mn + O2 = 2MnO nC(for CO2) = (1/3) * 72.917 = 24.306 kg-atom Weight of Mn in Pig iron = 0.01 * 25000 = 250 kg nO2= nC(for CO2) = 24.306 kg-atom nMn= 250/55 = 4.545 kg-atom nO2= 4.545/2 = 2.273 kg-mole Oxidation step 4: 4P + 5O2 = 2P2O5 Weight of P in iron = 0.025 * 25000 = 625 kg nP=625/31 = 20.161 kg-atom nO2=(5/4) * nP=25.201 kg-mole

Flux CaO 25 tons Pig iron Bessemer converter 91% Fe 3.5% C Air 2% Si 500 m3/min 1% Mn 2.5% P Calculate the volume of air required for the operation Basis 25 tons of pig iron

Flue gas CO, CO2, N2 Slag Fe2O3, P2O5, SiO2 MnO, 35% CaO Steel

Total O2 used during Si, Mn, C, P oxidation = 93.94 kg-mole Considering small amount of Fe oxidizing in all steps: nFe = 1137.5/56 = 20.312 kg-atom 2Fe + 3/2 O2 = Fe2O3 nO2 = (¾)*nFe = (¾)*20.312 = 15.23 kg-mole

Total O2 used = 2.273 + 17.857 + 24.306 + 24.306 + 25.201 + 15.23 = 109.17 kg-mole O2 Volume of air required = (109.17/0.21) * 22.4 = 11644.8 m3/25 ton of Pig iron Total blowing time = 11644.8/500 = 23.29 minutes

Flux CaO

Flue gas CO, CO2, N2

25 tons Pig iron Bessemer converter 91% Fe 3.5% C Air 2% Si 500 m3/min 1% Mn 2.5% P Calculate the durations of each blowing period

Slag Fe2O3, P2O5, SiO2 MnO, 35% CaO Steel

Total O2 used = 2.273 + 17.857 + 24.306 + 24.306 + 25.201 + 15.23 = 109.17 kg-mole O2 Volume of air required for period 1= (2.273/0.21) * 22.4 = 242.45 m3/500 kg Si Volume of air required for period 2= (17.857/0.21) * 22.4 = 1904.75 m3/250 kg Mn Volume of air required for period 3= (48.162/0.21) * 22.4 = 5137.3 m3/878 kg C Volume of air required for period 4= (25.201/0.21) * 22.4 = 2688.1 m3/625 kg P Volume of air distributed in all periods= (15.23/0.21) * 22.4 = 1624.5 m3/1138 kg Fe Total blowing time = 11644.8/500 = 23.29 minutes FeFe2O3 Period 1 Period 2 Period 3 Period 4 SiSiO2 MnMnO 0

4.43 4.99

CCO, CO2 Time (min)

PP2O5 17.04

23.29

Flux CaO 25 tons Pig iron Bessemer converter 91% Fe 3.5% C Air 2% Si 500 m3/min 1% Mn 2.5% P Calculate the weight of CaO added to the converter Si + O2 = SiO2 Weight of Si in Pig iron = 0.02 * 25000 = 500 kg nSi= 500/28 = 17.857 kg-atom nSiO2= 17.857 kg-mole Weight of SiO2 in slag = 17.857*60 = 1071.4 kg

Flue gas CO, CO2, N2 Slag Fe2O3, P2O5, SiO2 MnO, 35% CaO Steel

4P + 5O2 = 2P2O5 Weight of P in iron = 0.025 * 25000 = 625 kg nP=625/31 = 20.161 kg-atom nP2O5= 20.161/2 = 10.081 kg-mole Weight of P2O5 in slag = 17.081*142 = 1431.5 kg

2Mn + O2 = 2MnO Weight of Mn in Pig iron = 0.01 * 25000 = 250 kg 2Fe + 3/2 O2 = Fe2O3 nMn= 250/55 = 4.545 kg-atom nFe = 1137.5/56 = 20.312 kg-atom nMnO= 4.545 kg-mole nFe2O3 = 20.312/2 = 10.156 kg-mole Weight of MnO in slag = 4.545*71 = 322.7 kg Weight of Fe2O3 in slag = 10.156*160 = 1625kg MnO + SiO2 + P2O5 + Fe2O3 = 4450.6 kg, 35% CaO= (4450.6/0.65)*0.35 = 2396.5 kg CaO in slag

Flux CaO 25 tons Pig iron 91% Fe 3.5% C Air 2% Si 500 m3/min 1% Mn 2.5% P

Flue gas CO, CO2, N2

Bessemer converter

Steel

Calculate the weight and composition of the slag Component SiO2 MnO P2O5 Fe2O3 CaO Total

Weight (kg) 1071.4 322.7 1431.5 1625 2396.5 6848

Slag Fe2O3, P2O5, SiO2 MnO, 35% CaO

Weight % 15.64% 4.71 % 20.90% 23.75% 35.00% 100.00%

Flux CaO

Flue gas CO, CO2, N2

25 tons Pig iron Bessemer converter Slag 91% Fe Fe2O3, P2O5, SiO2 3.5% C Air MnO, 35% CaO 2% Si 500 m3/min 1% Mn Steel 2.5% P Calculate the weight of CaO added to the converter in each blowing period Total CaO added as flux = 2396.5 kg CaO

Period 1

Period 2

FeFe2O3 Period 3

SiSiO2 MnMnO 0

4.43 4.99

Period 4

CCO, CO2 Time (min)

CaO consumed each minute = 2396.5/23.29 = 102.9 kg CaO consumed in period 1 = 102.9 * 4.43 = 455.8 kg CaO consumed in period 2 = 102.9 * 0.56 = 58.02 kg CaO consumed in period 3 = 102.9 * 11.95 = 1229.4 kg CaO consumed in period 4 = 102.9 * 6.25 = 643.3 kg

PP2O5 17.04

23.29

Copper converting Liquid matte from the smelting process is oxidized in a bessemer or basic oxygen furnace by blowing air or oxygen The difference between steel converting and copper converting is that the value mineral Cu2S is oxidized as well in the latter process

Cu2 S + O2 = Cu l + SO2 FeS2 in the matte is oxidized initially due to its higher oxidation free energy Excess S in the matte may also oxidize preferentially prior to reduction of copper Blowing, fluxing and slagging may be done periodically due to convenience Flux is commonly added in batches due to the high amount of charge material and the limited space of furnaces The amount of flux batches and blowing rate affects the time taken to produce slag in periods SO2 on the surface of the copper evaporate and form blisters on the solidifying copper Blister copper purity is around 99% and electrolysis treatment is needed to obtain pure copper Converted blister copper is considered as 100% pure for convenience in material balance

Copper converter analysis 40 tons of matte carrying 34% Cu is charged in a bessemer converter The flux is added in batches of 3000 kg, the converter is blown after each addition to obtain slag of the given composition Blister copper is produced after the removal of slags formed using partially added fluxes Air is blown at a rate of 100 m3/minute

Flux

Slag

Matte

Bessemer converter

Blister copper

Air

Rational Analysis wt% Material

FeO

CaO

SiO2

Al2O3

Slag

59

8

32

1

?

75

?

Flux

Cu2S

FeS2

2

5

Air is blown at a rate of 100 m3/minute 3000 kg Flux 75% SiO2 2% Cu2S 5% FeS2 Bessemer converter 40 tons Matte 34% Cu Air 100 m3/min Calculate the time of each partial blow Cu2S in matte = 40000 * 0.34 * (160/128) = 17000 kg, Let X be the weight of slag SiO2 balance: 75% * 3000 = 0.32 X X = 7031 kg

87.725 0.21

Blister copper

FeS=23000 kg

Let Y be the weight of FeS oxidized in one blow Fe balance: (56/88) * Y + 5% * 3000 * (56/120) = 59% * 7031 * (56/72) Y = 4963 kg = 56.4 kg-moles

Oxygen required for 1 blow: FeS + 3/2O2 = FeO + SO2 O2 required = 56.4 * (3/2) = 84.6 kg-moles Total O2 required = 87.725 kg-moles Time for 1 blow =

Slag 59% FeO, 8 CaO 32% SiO2, 1% Al2O3

∗

22.4 100

FeS2 + 5/2O2 = FeO + 2SO2 O2 required = 1.25 * (5/2) = 3.125 kg-moles

= 93.57 minutes

Air is blown at a rate of 100 m3/minute 3000 kg Flux 75% SiO2 2% Cu2S 5% FeS2 Bessemer converter 40 tons Matte 17 tons Cu2S 23 tons FeS

7031 kg Slag 59% FeO, 8 CaO 32% SiO2, 1% Al2O3 Blister copper

Air 100 m3/min

Calculate the number of partial blows and the weight of flux to be added for the last partial blow to completely remove FeO in the slag 87.725

22.4

Time for 1 blow = 0.21 ∗ 100 = 93.57 minutes The weight of FeS oxidized in one blow = 4963 kg = 56.4 kg-moles Number of partial blows = 23000/4963 = 4.63 ≈ 5 FeS oxidized in the 5th blow = 23000 – (4*4963) = 3148 kg Let Z be the amount of flux batch added in the last period O2 balance: (3148/88) *(3/2) + (0.05Z /120) *(5/2) = 87.725 * 3148/4963 Z= 1903 kg

Air is blown at a rate of 100 m3/minute 3000 kg Flux 75% SiO2 2% Cu2S 5% FeS2 Bessemer converter 40 tons Matte 17 tons Cu2S 23 tons FeS

7031 kg Slag 59% FeO, 8 CaO 32% SiO2, 1% Al2O3 Blister copper

Air 100 m3/min

Calculate the total time for blowing the charge to convert to blister copper Blow Period (min) 1 93.57 2 93.57 3 93.57 4 93.57 5 3148/4963 * 93.57 = 59.40 minutes Total time to remove Fe in the matte and flux completely = 431.68 minutes Cu2 S + O2 = Cu l + SO2 Total Cu2S = 17000/160 = 108 kg-moles, Total O2 required = 108 kg-moles Total air required = (108/0.21) * 22.4 = 11520 m3, Time required to convert Cu = 115.2 minutes Total time of operation = 546.88 minutes